Inequality challenge proof day 2

I.. $A,B,C$ are angles of a triangle. Prove that

$\sum_{\mathrm{cyc}}\cos A\leq\sum_{\mathrm{cyc}}\sin\left(\frac{A}{2}\right)$

ii. Consider $a,b,c>0, abc=1$ . Prove that

$\sum_{\mathrm{cyc}} \frac{a^{2}+b^{2}}{a^{8}+b^{8}}\leq a^{3}+b^{3}+c^{3}$

iii. For $a,b>0$ , prove that $(1+a)^{8}+(1+b)^{8}\geq 128ab(a+b)^{2}$

iv. $p,q,r$ are distinct prime numbers such that $rp^{3}+p^{2}+p=2rq^{2}+q^{2}+q$ . Find all possible values of $pqr$ .

v. If $a,b$ are integers, then prove that at least one of these expressions is an integer: $\frac{a^{2}+b}{b^{2}-a},\frac{b^{2}+a}{a^{2}-b}$ .

vi. $\sum_{cyclic}^{a,b,c}\frac{(b+c)(a^{4}-(bc)^{2})}{ab+2bc+ac}\geq0$ .

#Algebra #RMO #NMTC

Easy Math Editor

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Comments

Q1) $\cos A + \cos B = 2 \cos \left(\dfrac{A+B}{2} \right) \cos \left(\dfrac{A-B}{2} \right)$

Since $A,B,C$ are angles of triangle. This gives us that $A+B+C= \pi \implies A+B = \pi - C$ . We also have $\cos \left(\dfrac{A-B}{2} \right) \leq 1$

So, $\begin{aligned} \cos A + \cos B &= 2 \cos \left(\dfrac{\pi-C}{2} \right) \cos \left(\dfrac{A-B}{2} \right) \\ &= 2 \sin \dfrac{C}{2} \cos \left(\dfrac{A-B}{2} \right) \\ & \leq 2\sin \dfrac{C}{2} \end{aligned}$

Similarly we get $\cos B + \cos C \leq 2\sin \dfrac{A}{2}$ and $\cos C + \cos A \leq 2\sin \dfrac{B}{2}$ .

Adding all these we get,

$\sum_{cyc} \cos A \leq \sum_{cyc} \sin \dfrac{A}{2}$

And equality holds when $A=B=C = \dfrac{\pi}{3}$ .

Surya Prakash - 5 years, 6 months ago

What does question 4 ask for?

Jake Lai - 5 years, 7 months ago

Its correct now.

Shivam Jadhav - 5 years, 7 months ago

q3)

$(1+a)^8+(1+b)^8\ge (2\sqrt{a})^8+(2\sqrt{b})^8=256(a^4+b^4)$ It remains to prove $256(a^4+b^4)\ge 128ab(a+b)^2$ which is obvious by expansion.

Daniel Liu - 5 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$